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From the memoryless property of Exponential Distribution function:
 
From the memoryless property of Exponential Distribution function:
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Suppose E1,λ and E1,μ are independent, then;
 
Suppose E1,λ and E1,μ are independent, then;
P[min{ E1,λ , E1,μ } > t] = P[E1,λ > t] . P[E1,μ } > t]
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        = eˉλt . eˉμt
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P [min{ E1,λ , E1,μ } > t] = P [E1,λ > t] . P [E1,μ } > t]
        = eˉ(λ + μ)t
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          = eˉλt . eˉμt
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          = eˉ(λ + μ)t
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which shows that minimum of E1,λ and E1,μ is exponentially distributed.
 
which shows that minimum of E1,λ and E1,μ is exponentially distributed.
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So,
 
So,
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E1, λ1+ λ2+ λ3+……. λn = min { E1,λ1, E1,λ2, E1,λ3, ……….., E1,λn }
 
E1, λ1+ λ2+ λ3+……. λn = min { E1,λ1, E1,λ2, E1,λ3, ……….., E1,λn }
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Here, if we put λ = 1, then;
 
Here, if we put λ = 1, then;
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E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n }
 
E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n }
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Revision as of 17:49, 6 October 2008

From the memoryless property of Exponential Distribution function: Suppose E1,λ and E1,μ are independent, then; P [min{ E1,λ , E1,μ } > t] = P [E1,λ > t] . P [E1,μ } > t] = eˉλt . eˉμt = eˉ(λ + μ)t which shows that minimum of E1,λ and E1,μ is exponentially distributed. So, E1, λ1+ λ2+ λ3+……. λn = min { E1,λ1, E1,λ2, E1,λ3, ……….., E1,λn } Here, if we put λ = 1, then; E1, 1+ 2+ 3+……. n = min { E1,1, E1,2, E1,3, ……….., E1,n }

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